Please help me out. Thank you

Question

A rock is thrown upward with a speed of 48 feet per second from the edge of a cliff 400 feet above the ground. What is the speed of the rock when it hits the ground? Use acceleration due to gravity as -32 feet per second squared and approximate your answer to 3 decimal places.

Answer 1

The rock will go up and down.

When it goes upward the gravity will be -32 feet per second square, and +32 feet per second square when it goes down.

When the rock goes up, it will reach the maximum height and the velocity at max height is 0

Vt = Vo - g.t (Vo is initial velocity; Vt is V after t sec)

0 = 48 - 32.t

32t = 48

t = 1.5 sec (the rock needs 1.5 sec to reach max height)

We need to find how high the rock goes upward

h = Vo.t - 0.5 g t^2

= 48 * 1.5 - 0.5 *32 * (1.5)^2

= 36 feet

The rock travel 36 feet above the cliff (400 ft above ground)

So the rock's height now is 436 feet above ground

The rock will free fall from 436 feet above ground (Vo = 0 ft/sec, h = 436ft)

We need to find the time for the rock to reach the ground, since the rock go down, gravity is 32ft/sec. Remember that gravity is a vector, the plus and minus sign are the direction.

h = Vo.t + 0.5 g t^2

436 = 0*t +0.5 *32 *t^2

436 = 16t^2 (divide both side with 16)

27.25 = t^2 (square root both sides)

t = 5.220 sec

Vt = Vo + g*t

= 0 + 32 * 5.220 sec

= 167.045 ft/sec (The velocity of the rock when it hits the ground)

Answer 2

So

we can break this problem into two parts

we can use y=y₀+(v_0y)*t+(1/2)*a*t² where we take up as positive and down as negative

v_oy=48ft/s a = -32 ft/s²

-400 = 48*t - (1/2)*32*t²

or t = 6.72 seconds

Next, use v² = v_o² + 2*a*y using a=-32 ft/s² vo=48 ft/s y=-400

v²=(48)²-2*32*(-400)

v^{2 =} 2304 + 25600 solve for v